The point on the curve 9y² = x³, where the normal to the curve makes equal intercepts with the axes is

Given curve 9y² = x³ ….(1)

Differentiate w.r.t. x,

Equation of normal:

∵ it makes equal intercepts with the axes

∴ slope of the normal = ±1

⇒ x² = ±6y

Squaring both the sides,

x⁴ = ± 36y²

From (1),

x= 0, 4

and

But the line making equal intercept cannot pass through origin.

So, the required points are